This question seems quite similar to the problem of parity domination: the condition that $S$ flips to its complement after updating is equivalent to the condition that $|N[v] \cap S|$ is odd for all $v \in V(G)$, where $N[v] = N(v) \cup \{v\}$. Such a set is called an odd dominating set, and it was proved by Sutner that every graph has such a set. In fact, Sutner's proof involved considering a similar CA. See, e.g., this paper for more about parity domination, or see here for Sutner's article.
Asking whether $G$ has a blinking assignment, then, is equivalent to asking whether $G$ has an odd dominating set whose complement is also an odd dominating set. Writing $d_S[v]$ and $d_{G-S}[v]$ for the size of $N[v] \cap S$ and $N[v] \cap (V(G)-S)$ respectively, we need that every $v \in V(G)$, the quantities $d_S[v]$ and $d_{G-S}[v]$ are both odd. Since $d_S[v] + d_{G-S}[v] = d(v) + 1$, this means every vertex of $G$ needs to have odd degree. On the other hand, if every vertex of $G$ has odd degree, then any odd dominating set should do the trick, and we're guaranteed that one exists.
So, I believe that $G$ should have a blinking assignment if and only if all its vertices have odd degree.